Complex Numbers Revision Notes — 9709 Paper 3

sub-topic	parts	marks
terminology: modulus, argument, conjugate	45	142
Cartesian arithmetic	34	139
loci	41	128
Argand diagram	41	122
conjugate-pair roots of polynomials	10	37
square roots	7	35
multiplying and dividing in polar form	13	31
geometrical effects of the operations	8	14

Complex numbers and Cartesian arithmetic

Syllabus requirement · §3.9

“

understand the idea of a complex number, recall the meaning of the terms real part, imaginary part, modulus, argument, conjugate, and use the fact that two complex numbers are equal if and only if both real and imaginary parts are equal; carry out operations of addition, subtraction, multiplication and division of two complex numbers expressed in Cartesian form x + iy. For calculations involving multiplication or division, full details of the working should be shown.

”

Where this comes from

You have met this problem before without being told it had a solution. Solve $x^2 - 4x + 13 = 0$ with the quadratic formula and you get

x = \frac{4 \pm \sqrt{16-52}}{2} = \frac{4 \pm \sqrt{-36}}{2}

At AS you stopped here and wrote "no real roots", because no real number squares to give $-36$ . Complex numbers pick the sentence up again. Invent a number whose square is $-1$ and call it $\mathrm{i}$ :

\mathrm{i}^2 = -1 \qquad\text{so}\qquad \sqrt{-1} = \mathrm{i}

Then $\sqrt{-36} = \sqrt{36}\times\sqrt{-1} = 6\mathrm{i}$ , and the equation has the two roots

x = \frac{4 \pm 6\mathrm{i}}{2} = 2 \pm 3\mathrm{i}

That sounds like cheating, and it is worth saying clearly why it is not. Mathematics has done this before: negative numbers were invented so that $3-5$ had an answer, and $\sqrt2$ so that $x^2=2$ did. Each time the test is the same — does the new number break any arithmetic? It does not. Every ordinary rule of algebra (expanding brackets, collecting like terms, factorising) still holds. The only new rule is: whenever $\mathrm{i}^2$ appears, replace it by $-1$ .

The name "imaginary" is a historical insult that stuck. There is nothing less real about $\mathrm{i}$ than about $-1$ .

The shape of a complex number

Because $\mathrm{i}^2$ can always be replaced by $-1$ , no higher powers of $\mathrm{i}$ ever survive. Everything collapses to one real part plus one multiple of $\mathrm{i}$ :

z = x + \mathrm{i}y, \qquad x \text{ and } y \text{ both real}

This is called the Cartesian form of $z$ (the same word as Cartesian coordinates, and for the reason you will see in the Argand diagram section below: $x$ and $y$ will become coordinates).

$x$ is the real part, written $\operatorname{Re}z$ .
$y$ is the imaginary part, written $\operatorname{Im}z$ .

Read that second definition carefully, because it is the commonest slip in the topic: for $z = 3+2\mathrm{i}$ the imaginary part is $\mathbf{2}$ , not $2\mathrm{i}$ . Both $\operatorname{Re}z$ and $\operatorname{Im}z$ are ordinary real numbers.

Two special cases have names that questions use:

If $y = 0$ , the number is just real. So every real number is already a complex number — $5 = 5 + 0\mathrm{i}$ .
If $x = 0$ , the number is called purely imaginary, like $4\mathrm{i}$ . "Given that $z$ is purely imaginary" means "set $\operatorname{Re}z = 0$ ", and "given that $z$ is real" means "set $\operatorname{Im}z = 0$ ".

The letter $z$ is traditional for a complex number, just as $x$ is for a real one; $w$ , $u$ and $v$ are also common.

The powers of i go round in fours

Build them up one multiplication at a time:

\mathrm{i}^1 = \mathrm{i}, \qquad \mathrm{i}^2 = -1, \qquad \mathrm{i}^3 = \mathrm{i}^2\times\mathrm{i} = -\mathrm{i}, \qquad \mathrm{i}^4 = \mathrm{i}^2\times\mathrm{i}^2 = (-1)(-1) = 1

After four the pattern repeats, because multiplying by $\mathrm{i}^4 = 1$ changes nothing. So to simplify a high power, divide the index by $4$ and keep only the remainder: $\mathrm{i}^{11} = \mathrm{i}^{8}\times\mathrm{i}^{3} = 1\times(-\mathrm{i}) = -\mathrm{i}$ .

This cycle is the first hint of the geometry later in the note: multiplying by $\mathrm{i}$ turns a number through a quarter turn, and four quarter turns bring it back to where it started.

Adding, subtracting and multiplying

Addition and subtraction work part by part — real with real, imaginary with imaginary — exactly like collecting like terms in $x$ and $y$ :

(3+2\mathrm{i}) + (1-4\mathrm{i}) = (3+1) + (2-4)\mathrm{i} = 4 - 2\mathrm{i}

(3+2\mathrm{i}) - (1-4\mathrm{i}) = (3-1) + \big(2-(-4)\big)\mathrm{i} = 2 + 6\mathrm{i}

Multiplication is ordinary bracket expansion. Nothing is special until the $\mathrm{i}^2$ term appears, and then it is replaced by $-1$ :

(3+2\mathrm{i})(1-4\mathrm{i}) = \underbrace{3 - 12\mathrm{i} + 2\mathrm{i}}_{\text{three ordinary terms}} \;\underbrace{-\; 8\mathrm{i}^2}_{-8\times(-1)\,=\,+8} = 11 - 10\mathrm{i}

Notice what the $-8\mathrm{i}^2$ did: it became $+8$ and moved into the real part. That migration — an imaginary times an imaginary landing in the real part — is the one genuinely new feature of complex multiplication, and it is where most sign errors come from. Write the $\mathrm{i}^2$ term out, then replace it, as two separate steps.

The four operations, and powers, on invented numbers

Given $z = 2 - \mathrm{i}$ and $w = 3 + 4\mathrm{i}$ , find in the form $x + \mathrm{i}y$ : (a) $z + w$ , (b) $w - 2z$ , (c) $zw$ , (d) $z^2$ and $z^3$ .

Show full working

1
(a) Add the real parts, then add the imaginary parts: $z + w = (2 + 3) + (-1 + 4)\mathrm{i} = 5 + 3\mathrm{i}$
2
(b) First find $2z$ by multiplying both parts by $2$ : $2z = 2(2 - \mathrm{i}) = 4 - 2\mathrm{i}$
A real multiplier scales the real part and the imaginary part alike — the same as 2(x − y) = 2x − 2y.
3
Now subtract part by part, taking care with the double negative in the imaginary part: $w - 2z = (3 - 4) + \big(4 - (-2)\big)\mathrm{i} = -1 + 6\mathrm{i}$
4
(c) Expand $(2 - \mathrm{i})(3 + 4\mathrm{i})$ term by term, first times first, and so on: $zw = 2(3) + 2(4\mathrm{i}) + (-\mathrm{i})(3) + (-\mathrm{i})(4\mathrm{i}) = 6 + 8\mathrm{i} - 3\mathrm{i} - 4\mathrm{i}^2$
5
Replace $\mathrm{i}^2$ by $-1$ , so $-4\mathrm{i}^2 = -4(-1) = +4$ : $zw = 6 + 8\mathrm{i} - 3\mathrm{i} + 4$
This is the step examiners look for — the M1 for ‘use i² = −1’ is awarded on seeing it.
6
Collect real and imaginary parts: $zw = (6 + 4) + (8 - 3)\mathrm{i} = 10 + 5\mathrm{i}$
7
(d) For $z^2$ , expand $(2 - \mathrm{i})^2$ like $(a - b)^2 = a^2 - 2ab + b^2$ : $z^2 = 2^2 - 2(2)(\mathrm{i}) + \mathrm{i}^2 = 4 - 4\mathrm{i} + \mathrm{i}^2 = 4 - 4\mathrm{i} - 1 = 3 - 4\mathrm{i}$
8
For $z^3$ , do not start again — multiply the $z^2$ you already have by $z$ once more: $z^3 = z^2 \times z = (3 - 4\mathrm{i})(2 - \mathrm{i}) = 6 - 3\mathrm{i} - 8\mathrm{i} + 4\mathrm{i}^2$
Building powers one multiplication at a time keeps every line short. It is exactly how the mark schemes lay out z² and z³ when a root is substituted into a cubic.
9
Replace $4\mathrm{i}^2$ by $-4$ and collect: $z^3 = 6 - 4 - 11\mathrm{i} = 2 - 11\mathrm{i}$

Answer

(a) $5 + 3\mathrm{i}$ (b) $-1 + 6\mathrm{i}$ (c) $10 + 5\mathrm{i}$ (d) $z^2 = 3 - 4\mathrm{i}$ , $\;z^3 = 2 - 11\mathrm{i}$

Powers of a complex number come up constantly — whenever a complex root is substituted into a cubic or quartic. Always build z³ as z² × z, and z⁴ as z² × z².

Equal means equal in both parts

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal:

a + \mathrm{i}b = c + \mathrm{i}d \quad\Longleftrightarrow\quad a = c \;\text{ and }\; b = d \qquad (a, b, c, d \text{ real})

Why must this be true? Rearranging gives $a - c = \mathrm{i}(d - b)$ . The left side is real. If $d - b$ were not zero, the right side would be a non-zero multiple of $\mathrm{i}$ , and no real number squares to a negative — so $d - b = 0$ , and then $a - c = 0$ as well.

This single fact turns one complex equation into two real equations. It is the engine behind unknown-constant problems, equations involving $z^*$ , square roots and much else. Whenever a question contains unknown real numbers, equating real and imaginary parts is almost always the intended route.

Equating parts to find two real unknowns

Find the real numbers $x$ and $y$ such that $(x + \mathrm{i}y)(1 + 2\mathrm{i}) = 7 + 4\mathrm{i}$ .

Show full working

1
Expand the left side as an ordinary product: $(x + \mathrm{i}y)(1 + 2\mathrm{i}) = x + 2\mathrm{i}x + \mathrm{i}y + 2\mathrm{i}^2y$
2
Replace $\mathrm{i}^2$ by $-1$ , so $2\mathrm{i}^2y = -2y$ , and group into real part plus imaginary part: $(x - 2y) + \mathrm{i}(2x + y)$
Group before equating. Every term without an i goes in the first bracket; every term with an i goes in the second.
3
The equation now reads $(x - 2y) + \mathrm{i}(2x + y) = 7 + 4\mathrm{i}$ . Equate real parts and equate imaginary parts: $x - 2y = 7 \qquad\text{and}\qquad 2x + y = 4$
4
Solve the pair. Double the second equation, $4x + 2y = 8$ , and add it to the first to eliminate $y$ : $5x = 15 \;\Longrightarrow\; x = 3$
5
Substitute back into $2x + y = 4$ : $6 + y = 4 \;\Longrightarrow\; y = -2$
6
Check by multiplying out: $(3 - 2\mathrm{i})(1 + 2\mathrm{i}) = 3 + 6\mathrm{i} - 2\mathrm{i} - 4\mathrm{i}^2 = 3 + 4\mathrm{i} + 4 = 7 + 4\mathrm{i}$ ✓

Answer

$x = 3$ , $y = -2$

One complex equation = two real equations. Once you have written the left side as (real) + i(real), the rest is GCSE simultaneous equations.

The conjugate, and why it makes division possible

Division is the one operation with no obvious meaning. $\dfrac{4+7\mathrm{i}}{2+3\mathrm{i}}$ is not in the form $x+\mathrm{i}y$ , and there is no way to read off its real part while an $\mathrm{i}$ sits in the denominator.

The complex conjugate solves this. The conjugate of $z = x+\mathrm{i}y$ is written $z^*$ and is the same number with the sign of the imaginary part flipped:

z = x + \mathrm{i}y \qquad\Longrightarrow\qquad z^* = x - \mathrm{i}y

So $(3 + 2\mathrm{i})^* = 3 - 2\mathrm{i}$ , $(-1 - 5\mathrm{i})^* = -1 + 5\mathrm{i}$ , and a real number is its own conjugate. The crucial property is that a number times its own conjugate is real:

z z^* = (x+\mathrm{i}y)(x-\mathrm{i}y) = x^2 - \mathrm{i}xy + \mathrm{i}xy - \mathrm{i}^2y^2

The two middle terms cancel, and $-\mathrm{i}^2y^2 = -(-1)y^2 = +y^2$ , leaving

z z^* = x^2+y^2

It is the difference of two squares, $(x+\mathrm{i}y)(x-\mathrm{i}y) = x^2 - (\mathrm{i}y)^2$ , with the minus sign turned into a plus by $\mathrm{i}^2 = -1$ .

So to divide: multiply the top and the bottom by the conjugate of the bottom. That multiplies the fraction by $1$ , so its value is unchanged, but the denominator becomes real and the fraction can then be split into its two parts.

Dividing, step by step

1
Write down the conjugate of the denominator. For $\dfrac{4+7\mathrm{i}}{2+3\mathrm{i}}$ the denominator is $2+3\mathrm{i}$ , so its conjugate is $2-3\mathrm{i}$ .
The conjugate of the denominator, never of the numerator — the aim is to make the bottom real.
2
Multiply top and bottom by it: $\frac{4+7\mathrm{i}}{2+3\mathrm{i}} \times \frac{2-3\mathrm{i}}{2-3\mathrm{i}}$
(2 − 3i)/(2 − 3i) equals 1, so the value is unchanged — only its appearance changes.
3
Expand the denominator first — it must come out real: $(2+3\mathrm{i})(2-3\mathrm{i}) = 2^2 + 3^2 = 13$
A number times its conjugate is always (real part)² + (imaginary part)². If an i survives here, you used the wrong conjugate.
4
Expand the numerator as an ordinary product: $(4+7\mathrm{i})(2-3\mathrm{i}) = 8 - 12\mathrm{i} + 14\mathrm{i} - 21\mathrm{i}^2$
5
Replace $\mathrm{i}^2$ by $-1$ and collect: $8 + 2\mathrm{i} + 21 = 29 + 2\mathrm{i}$
6
Split into real and imaginary parts over the real denominator: $\frac{29+2\mathrm{i}}{13} = \frac{29}{13} + \frac{2}{13}\mathrm{i}$
“In the form x + iy” means the answer must be split like this. One mark scheme gives A0 for an answer left as the single fraction (3 + 4i)/5.

You have done this before

The move is exactly rationalising a surd denominator, and for the same reason:

\frac{5}{2+\sqrt3} \times \frac{2-\sqrt3}{2-\sqrt3} = \frac{5(2-\sqrt3)}{4-3} \qquad\qquad \frac{5}{2+3\mathrm{i}} \times \frac{2-3\mathrm{i}}{2-3\mathrm{i}} = \frac{5(2-3\mathrm{i})}{4+9}

Both use a difference of two squares to clear the denominator. The only difference is that $\left(\sqrt3\right)^2 = +3$ while $\left(3\mathrm{i}\right)^2 = 9\mathrm{i}^2 = -9$ , which is why one denominator subtracts and the other adds.

Show the working

The syllabus says it outright: "For calculations involving multiplication or division, full details of the working should be shown."

A calculator will do complex arithmetic, and the mark schemes say it plainly: "Correct answer with no working scores 0/3." Write the conjugate you are multiplying by, the expanded numerator and the real denominator, every time.

Dividing when the numbers contain a letter

9709/31 O/N 2023 Q4(a)3 marks

The complex number $u$ is defined by $u = \dfrac{3 + 2\mathrm{i}}{a - 5\mathrm{i}}$ , where $a$ is real.

Express $u$ in the Cartesian form $x + \mathrm{i}y$ , where $x$ and $y$ are in terms of $a$ .

Show full working

1
The denominator is $a - 5\mathrm{i}$ . Its real part is $a$ and its imaginary part is $-5$ , so its conjugate is $a + 5\mathrm{i}$ . Multiply top and bottom by it: $u = \frac{3 + 2\mathrm{i}}{a - 5\mathrm{i}} \times \frac{a + 5\mathrm{i}}{a + 5\mathrm{i}}$
The letter a changes nothing: it is just a real number whose value you don't know yet, so it stays in the real part.
2
Denominator: a number times its conjugate is (real part) $^2$ + (imaginary part) $^2$ : $(a - 5\mathrm{i})(a + 5\mathrm{i}) = a^2 + 25$
3
Numerator: expand term by term: $(3 + 2\mathrm{i})(a + 5\mathrm{i}) = 3a + 15\mathrm{i} + 2a\mathrm{i} + 10\mathrm{i}^2$
4
Replace $10\mathrm{i}^2$ by $-10$ , then put real terms together and $\mathrm{i}$ terms together: $3a - 10 + (15 + 2a)\mathrm{i}$
5
Divide each part by the real denominator: $u = \frac{3a - 10}{a^2 + 25} + \frac{2a + 15}{a^2 + 25}\,\mathrm{i}$
Now x = (3a − 10)/(a² + 25) and y = (2a + 15)/(a² + 25) can be read off — and a later part of this question uses exactly those two expressions.

Answer

$u = \dfrac{3a-10}{a^2+25} + \dfrac{2a+15}{a^2+25}\,\mathrm{i}$

Bracket the imaginary part's coefficient when it has more than one term: (2a + 15)i, not 2a + 15i.

Equations in $z$

An equation that contains $z$ but not $z^*$ can be solved exactly as in ordinary algebra: collect the $z$ terms, factorise, divide. The only complex step is the final division, done with the conjugate.

For example, to solve $(2 + \mathrm{i})z = 7 - \mathrm{i}$ :

z = \frac{7 - \mathrm{i}}{2 + \mathrm{i}} \times \frac{2 - \mathrm{i}}{2 - \mathrm{i}} = \frac{14 - 7\mathrm{i} - 2\mathrm{i} + \mathrm{i}^2}{4 + 1} = \frac{13 - 9\mathrm{i}}{5} = \frac{13}{5} - \frac{9}{5}\mathrm{i}

The alternative, which the mark schemes also accept, is to put $z = x + \mathrm{i}y$ at the start and equate parts. That always works, but it creates two unknowns where one would do.

Two unknown real constants

9709/32 O/N 2023 Q87 marks

It is given that $\dfrac{2 + 3a\mathrm{i}}{a + 2\mathrm{i}} = \lambda(2 - \mathrm{i})$ , where $a$ and $\lambda$ are real constants.

(a) Show that $3a^2 + 4a - 4 = 0$ .

(b) Hence find the possible values of $a$ and the corresponding values of $\lambda$ .

Show full working

1
(a) Clear the fraction by multiplying both sides by $a + 2\mathrm{i}$ : $2 + 3a\mathrm{i} = \lambda(2 - \mathrm{i})(a + 2\mathrm{i})$
Cross-multiplying is quicker here than dividing by the conjugate, because it avoids a fraction on the left altogether.
2
Expand the product of the two brackets on the right: $(2 - \mathrm{i})(a + 2\mathrm{i}) = 2a + 4\mathrm{i} - a\mathrm{i} - 2\mathrm{i}^2$
3
Replace $-2\mathrm{i}^2$ by $+2$ and group: $(2a + 2) + (4 - a)\mathrm{i}$
4
So the equation is $\;2 + 3a\mathrm{i} = \lambda(2a + 2) + \lambda(4 - a)\mathrm{i}$ . Both $a$ and $\lambda$ are real, so equate real parts and equate imaginary parts: $2 = \lambda(2a + 2) \qquad\text{and}\qquad 3a = \lambda(4 - a)$
λ and a are real, so λ(2a + 2) really is the whole real part of the right side, and λ(4 − a) the whole imaginary part.
5
Eliminate $\lambda$ . From the first equation: $\lambda = \frac{2}{2a + 2} = \frac{1}{a + 1}$
6
Substitute into the second: $3a = \frac{4 - a}{a + 1}$
7
Multiply both sides by $a + 1$ : $3a(a + 1) = 4 - a \;\Longrightarrow\; 3a^2 + 3a = 4 - a$
8
Bring everything to one side: $3a^2 + 4a - 4 = 0 \quad\blacksquare$
A ‘show that’ is only complete when the given equation appears exactly, from correct working.
9
(b) Factorise the quadratic (the product $3\times(-4) = -12$ splits as $6$ and $-2$ ): $(3a - 2)(a + 2) = 0 \;\Longrightarrow\; a = \tfrac23 \;\text{ or }\; a = -2$
10
Find each $\lambda$ from $\lambda = \dfrac{1}{a + 1}$ . For $a = \tfrac23$ : $\lambda = \frac{1}{\tfrac53} = \frac35$
11
For $a = -2$ : $\lambda = \frac{1}{-2 + 1} = -1$

Answer

(b) $a = \tfrac23,\ \lambda = \tfrac35$ or $a = -2,\ \lambda = -1$

Unknown real constants inside a complex equation: expand, write each side as (real) + i(real), equate parts, then solve the real simultaneous equations. The values must come in matching pairs.

Equations containing both $z$ and $z^*$

A recurring Paper 3 question puts $z$ and $z^*$ in the same equation:

\frac{5z}{2-\mathrm{i}} - zz^* + 20 + 8\mathrm{i} = 0

This cannot be solved by treating $z$ as a single unknown, because $z$ and $z^*$ are different numbers — you cannot collect them into one term or cancel one against the other. The only route is to go back to the two real unknowns: put $z = x + \mathrm{i}y$ and $z^* = x - \mathrm{i}y$ , and let equating parts give two real equations.

The step that keeps it manageable is recognising $zz^* = x^2 + y^2$ , which is real. So a $zz^*$ term only ever contributes to the real part.

Solving an equation involving z*

1
Clear any fraction first — either multiply the whole equation by the denominator, or multiply that fraction top and bottom by the conjugate of its denominator.
The mark scheme's first method mark is for exactly this, before any substitution.
2
Substitute $z = x+\mathrm{i}y$ and $z^* = x - \mathrm{i}y$ everywhere, writing $zz^*$ straight away as $x^2+y^2$ .
3
Expand, replacing each $\mathrm{i}^2$ by $-1$ , and gather everything into the form (real) $+ \,\mathrm{i}\,$ (real) $= 0$ .
4
Set the real part and the imaginary part to zero separately. You now have two real simultaneous equations.
5
Solve them. Usually one equation is linear: rearrange it for one letter and substitute into the other.
Substituting the linear equation into the quadratic one leaves a single quadratic in one variable.
6
Give every solution in the form $x+\mathrm{i}y$ — there are usually two — and apply any condition the question states (such as $\operatorname{Re} z \leqslant 0$ ).

An equation with z and z* together

9709/33 O/N 2025 Q76 marks

Solve the equation $\dfrac{5z}{2-\mathrm{i}} - zz^* + 20 + 8\mathrm{i} = 0$ . Give your answers in the form $x+\mathrm{i}y$ , where $x$ and $y$ are real.

Show full working

1
Clear the fraction by multiplying it top and bottom by the conjugate of its denominator, $2+\mathrm{i}$ : $\frac{5z}{2-\mathrm{i}}\times\frac{2+\mathrm{i}}{2+\mathrm{i}} = \frac{5z(2+\mathrm{i})}{(2-\mathrm{i})(2+\mathrm{i})}$
2
The denominator is $2^2 + 1^2 = 5$ , which cancels the $5$ on top: $\frac{5z(2+\mathrm{i})}{5} = z(2+\mathrm{i})$
The question is built so this cancels exactly — a useful check that the conjugate was right.
3
The equation is now $\;z(2+\mathrm{i}) - zz^* + 20 + 8\mathrm{i} = 0$ . Substitute $z = x+\mathrm{i}y$ into the first term and expand: $(x+\mathrm{i}y)(2+\mathrm{i}) = 2x + \mathrm{i}x + 2\mathrm{i}y + \mathrm{i}^2 y$
4
Replace $\mathrm{i}^2y$ by $-y$ and group: $(2x - y) + \mathrm{i}(x+2y)$
5
The middle term is $zz^* = x^2+y^2$ , which is real. So the whole equation reads $(2x - y) + \mathrm{i}(x+2y) - (x^2 + y^2) + 20 + 8\mathrm{i} = 0$
6
Gather the real terms and the imaginary terms: $\left(2x - y - x^2 - y^2 + 20\right) + \mathrm{i}\left(x + 2y + 8\right) = 0$
7
Equate both parts to zero: $x + 2y + 8 = 0 \qquad\text{and}\qquad 2x - y - x^2 - y^2 + 20 = 0$
8
The first is linear. Rearrange it for $x$ : $x = -8 - 2y$
9
Substitute into the second equation: $2(-8-2y) - y - (-8-2y)^2 - y^2 + 20 = 0$
10
Expand each piece separately: $2(-8-2y) = -16 - 4y$ and $(-8-2y)^2 = (8 + 2y)^2 = 64 + 32y + 4y^2$ . So $-16 - 4y - y - 64 - 32y - 4y^2 - y^2 + 20 = 0$
(−8 − 2y)² is the same as (8 + 2y)², since squaring removes the overall minus sign. The minus in front of the bracket then applies to all three terms.
11
Collect: $y^2$ terms $-5y^2$ ; $y$ terms $-4y - y - 32y = -37y$ ; constants $-16 - 64 + 20 = -60$ : $-5y^2 - 37y - 60 = 0 \;\Longrightarrow\; 5y^2 + 37y + 60 = 0$
12
Factorise: $5\times60 = 300 = 12\times25$ and $12 + 25 = 37$ : $5y^2 + 12y + 25y + 60 = y(5y + 12) + 5(5y + 12) = (5y + 12)(y + 5) = 0$
13
So $y = -\tfrac{12}{5}$ or $y = -5$ .
14
Back-substitute into $x = -8 - 2y$ . For $y = -5$ : $x = -8 + 10 = 2$ . For $y = -\tfrac{12}{5}$ : $x = -8 + \tfrac{24}{5} = -\tfrac{16}{5}$ .

Answer

$z = 2 - 5\mathrm{i}$ or $z = -\tfrac{16}{5} - \tfrac{12}{5}\mathrm{i}$

Check one in the original: for z = 2 − 5i, zz* = 4 + 25 = 29 and z(2 + i) = 4 + 2i − 10i + 5 = 9 − 8i, so 9 − 8i − 29 + 20 + 8i = 0 ✓.

Your turn

A plain division first, then an equation in z alone, then equations with z* in them. In every one, write the i² term before replacing it by −1.

1
Express $\dfrac{3 + 4\mathrm{i}}{1 - 2\mathrm{i}}$ in the form $x + \mathrm{i}y$ , showing all your working.
Stuck? Show hint
Multiply top and bottom by the conjugate of 1 − 2i.
Show solution
1. 1
  The conjugate of the denominator $1 - 2\mathrm{i}$ is $1 + 2\mathrm{i}$ . Multiply top and bottom by it: $\frac{3 + 4\mathrm{i}}{1 - 2\mathrm{i}} \times \frac{1 + 2\mathrm{i}}{1 + 2\mathrm{i}}$
2. 2
  Denominator: $(1 - 2\mathrm{i})(1 + 2\mathrm{i}) = 1^2 + 2^2 = 5$ .
3. 3
  Numerator: $(3 + 4\mathrm{i})(1 + 2\mathrm{i}) = 3 + 6\mathrm{i} + 4\mathrm{i} + 8\mathrm{i}^2 = 3 + 10\mathrm{i} - 8 = -5 + 10\mathrm{i}$ .
4. 4
  Divide each part by $5$ : $\frac{-5 + 10\mathrm{i}}{5} = -1 + 2\mathrm{i}$
Answer
$-1 + 2\mathrm{i}$
29709/33 O/N 2024 Q45 marks
Find the complex number $z$ satisfying the equation $\frac{z - 3\mathrm{i}}{z + 3\mathrm{i}} = \frac{2 - 9\mathrm{i}}{5}.$ Give your answer in the form $x + \mathrm{i}y$ , where $x$ and $y$ are real.
Stuck? Show hint
There is no z* here, so treat z as a single unknown: cross-multiply, collect the z terms, factorise, then divide.
Show solution
1. 1
  Cross-multiply to clear both fractions: $5\left(z-3\mathrm{i}\right) = \left(2-9\mathrm{i}\right)\left(z+3\mathrm{i}\right)$
2. 2
  Expand the left side: $5z - 15\mathrm{i}$ . Expand the right side term by term: $\left(2-9\mathrm{i}\right)\left(z+3\mathrm{i}\right) = 2z + 6\mathrm{i} - 9\mathrm{i}z - 27\mathrm{i}^2$
  Keep z itself unexpanded — it is the unknown, handled like x in ordinary algebra.
3. 3
  Replace $-27\mathrm{i}^2$ by $+27$ , so the equation is $5z - 15\mathrm{i} = 2z + 6\mathrm{i} - 9\mathrm{i}z + 27$
4. 4
  Collect every term containing $z$ on the left and everything else on the right: $5z - 2z + 9\mathrm{i}z = 27 + 6\mathrm{i} + 15\mathrm{i}$
5. 5
  Factor out $z$ : $z\left(3+9\mathrm{i}\right) = 27+21\mathrm{i}$
6. 6
  Divide, and cancel the common factor $3$ : $z = \frac{27+21\mathrm{i}}{3+9\mathrm{i}} = \frac{9+7\mathrm{i}}{1+3\mathrm{i}}$
7. 7
  Multiply top and bottom by the conjugate $1 - 3\mathrm{i}$ . Denominator: $1^2 + 3^2 = 10$ . Numerator: $\left(9+7\mathrm{i}\right)\left(1-3\mathrm{i}\right) = 9 - 27\mathrm{i} + 7\mathrm{i} - 21\mathrm{i}^2 = 9 - 20\mathrm{i} + 21 = 30 - 20\mathrm{i}$
  The scheme needs to see this multiplication written out — just writing (9 + 7i)/(1 + 3i) = 3 − 2i scores nothing for the last two marks.
8. 8
  Divide by $10$ : $z = \frac{30-20\mathrm{i}}{10} = 3-2\mathrm{i}$
Answer
$z = 3-2\mathrm{i}$
39709/35 O/N 2025 Q55 marks
Find the complex numbers, $z$ , which satisfy the equation $zz^* + 5\mathrm{i}z + 2 - 10\mathrm{i} = 0.$ Give your answers in the form $x + \mathrm{i}y$ , where $x$ and $y$ are real.
Stuck? Show hint
Substitute z = x + iy, and remember zz* is just x² + y².
Show solution
1. 1
  Substitute $z=x+\mathrm{i}y$ . The first term is $zz^*=x^2+y^2$ .
2. 2
  The second term: $5\mathrm{i}z = 5\mathrm{i}\left(x+\mathrm{i}y\right) = 5\mathrm{i}x + 5\mathrm{i}^2y = 5\mathrm{i}x - 5y$
3. 3
  So the equation becomes $x^2 + y^2 + 5\mathrm{i}x - 5y + 2 - 10\mathrm{i} = 0$ . Group real and imaginary parts: $\left(x^2+y^2-5y+2\right) + \mathrm{i}\left(5x-10\right) = 0$
4. 4
  Equate both parts to zero: $x^2+y^2-5y+2=0 \qquad\text{and}\qquad 5x-10=0$
5. 5
  The imaginary-part equation gives $x=2$ at once.
6. 6
  Substitute $x = 2$ into the real-part equation: $4+y^2-5y+2=0 \;\Longrightarrow\; y^2-5y+6=0$
7. 7
  Factorise: $\left(y-2\right)\left(y-3\right)=0$ , so $y=2$ or $y=3$ .
Answer
$z = 2+2\mathrm{i}$ or $z = 2+3\mathrm{i}$
49709/32 F/M 2022 Q66 marks
Find the complex numbers $w$ which satisfy the equation $w^2 + 2\mathrm{i}w^* = 1$ and are such that $\operatorname{Re} w \leqslant 0$ . Give your answers in the form $x + \mathrm{i}y$ , where $x$ and $y$ are real.
Stuck? Show hint
Put w = x + iy and w* = x − iy. The imaginary-part equation factorises — each factor gives a case.
Show solution
1. 1
  Substitute: $\;(x + \mathrm{i}y)^2 + 2\mathrm{i}(x - \mathrm{i}y) = 1$ .
2. 2
  Expand the square: $(x + \mathrm{i}y)^2 = x^2 + 2\mathrm{i}xy + \mathrm{i}^2y^2 = x^2 - y^2 + 2\mathrm{i}xy$
3. 3
  Expand the second term: $2\mathrm{i}(x - \mathrm{i}y) = 2\mathrm{i}x - 2\mathrm{i}^2y = 2\mathrm{i}x + 2y$
4. 4
  Group, with the right side $1 = 1 + 0\mathrm{i}$ : $\left(x^2 - y^2 + 2y\right) + \mathrm{i}\left(2xy + 2x\right) = 1 + 0\mathrm{i}$
5. 5
  Equate parts: $x^2 - y^2 + 2y = 1 \qquad\text{and}\qquad 2xy + 2x = 0$
6. 6
  Factorise the imaginary-part equation: $2x(y + 1) = 0$ , so $x = 0$ or $y = -1$ .
  Don't divide by x — that would lose the case x = 0, which gives one of the two answers.
7. 7
  Case $x = 0$ : the real-part equation becomes $-y^2 + 2y = 1$ , i.e. $y^2 - 2y + 1 = 0$ , so $(y - 1)^2 = 0$ and $y = 1$ . This gives $w = \mathrm{i}$ , which has $\operatorname{Re} w = 0 \leqslant 0$ ✓.
8. 8
  Case $y = -1$ : the real-part equation becomes $x^2 - 1 - 2 = 1$ , so $x^2 = 4$ and $x = \pm 2$ .
9. 9
  The condition $\operatorname{Re} w \leqslant 0$ rejects $x = 2$ , leaving $x = -2$ and $w = -2 - \mathrm{i}$ .
  The scheme withholds the mark if 2 − i is also given — read the condition before writing the final list.
Answer
$w = \mathrm{i}$ or $w = -2 - \mathrm{i}$
59709/32 O/N 2021 Q36 marks
(a) Given the complex numbers $u = a + \mathrm{i}b$ and $w = c + \mathrm{i}d$ , where $a, b, c$ and $d$ are real, prove that $(u + w)^* = u^* + w^*$ .

(b) Solve the equation $(z + 2 + \mathrm{i})^* + (2 + \mathrm{i})z = 0$ , giving your answer in the form $x + \mathrm{i}y$ where $x$ and $y$ are real.
Stuck? Show hint
(a) Work out each side separately and compare. (b) Write z + 2 + i in Cartesian form first, then flip the sign of its imaginary part.
Show solution
1. 1
  (a) Left side: $u + w = (a + c) + \mathrm{i}(b + d)$ , so its conjugate is $(u + w)^* = (a + c) - \mathrm{i}(b + d)$
2. 2
  Right side: $u^* = a - \mathrm{i}b$ and $w^* = c - \mathrm{i}d$ , so $u^* + w^* = (a + c) - \mathrm{i}(b + d)$
3. 3
  The two sides are identical, so $(u + w)^* = u^* + w^*$ . $\blacksquare$
  A proof needs both sides worked out and an explicit statement that they match.
4. 4
  (b) Put $z = x + \mathrm{i}y$ . Then $z + 2 + \mathrm{i} = (x + 2) + \mathrm{i}(y + 1)$ , and flipping the sign of its imaginary part gives $(z + 2 + \mathrm{i})^* = (x + 2) - \mathrm{i}(y + 1)$
5. 5
  Expand the second term: $(2 + \mathrm{i})(x + \mathrm{i}y) = 2x + 2\mathrm{i}y + \mathrm{i}x + \mathrm{i}^2y = (2x - y) + \mathrm{i}(x + 2y)$
6. 6
  Add the two and group: $(x + 2 + 2x - y) + \mathrm{i}(-y - 1 + x + 2y) = (3x - y + 2) + \mathrm{i}(x + y - 1) = 0$
7. 7
  Equate both parts to zero: $3x - y + 2 = 0 \qquad\text{and}\qquad x + y - 1 = 0$
8. 8
  Add the two equations to eliminate $y$ : $4x + 1 = 0$ , so $x = -\tfrac14$ . Then $y = 1 - x = \tfrac54$ .
Answer
(b) $z = -\tfrac14 + \tfrac54\mathrm{i}$

Practise complex number arithmeticReal past-paper questions · Arithmetic of complex numbers in Cartesian form (x + iy)

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Can you do all of these?

Multiply out brackets, writing the i² term and then replacing it by −1
Divide by multiplying top and bottom by the conjugate of the denominator, showing the working
State Re z and Im z as real numbers
Equate real and imaginary parts to find unknown real constants
Solve an equation containing z and z* by substituting x + iy
Find the modulus, and the principal argument in any quadrant, exactly where possible
Convert between $x + \mathrm{i}y$ , $r(\cos\theta + \mathrm{i}\sin\theta)$ and $r\mathrm{e}^{\mathrm{i}\theta}$
Translate “is real”, “purely imaginary”, |z| = k and arg w = θ into equations
Multiply, divide and take powers in polar form, bringing the argument back into range
Name the geometric effect of each operation: translation, reflection, enlargement and rotation
Use moduli for sides and differences of arguments for angles in triangle proofs
Use conjugate pairs and sum/product to solve a cubic or quartic with real coefficients
Use the quadratic formula when a coefficient is complex — no conjugate pairs
Find both square roots, rejecting the negative value of x² and pairing the signs
Sketch circles, bisectors and half-lines with a scale, and shade the correct region
Write the inequalities for a given shaded region
Find greatest and least |z| and arg z, checking whether a tangent or a corner gives the extreme

Now do the questions

117 real Paper 3 parts from 2021–2025, sorted by difficulty, with mark schemes

Differential Equations

Complex Numbers

Complex numbers and Cartesian arithmetic

Where this comes from

The shape of a complex number

Adding, subtracting and multiplying

The conjugate, and why it makes division possible

Equations in $z$

Equations containing both $z$ and $z^*$

The Argand diagram: modulus, argument and polar form

Multiplying and dividing in polar form

Can you do all of these?

Complex numbers and Cartesian arithmetic

Where this comes from

The shape of a complex number

Adding, subtracting and multiplying

The conjugate, and why it makes division possible

Equations in zzz

Equations containing both zzz and z∗z^*z∗

The rest of this note

Can you do all of these?

Equations in $z$

Equations containing both $z$ and $z^*$